$f(-2)=7\,$, $~f\,^\prime(-2)=3\,$, $~f\,^{\prime\prime}(-2)=4\,$, and $~f\,^{\prime\prime\prime}(-2)=-8\,$. What are the first four nonzero terms of the Taylor series, centered at $x=-2$, of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $7+3(x-2)+2{{(x-2)}^{2}}-\frac{4}{3}{{(x-2)}^{3}}$ (Choice B) B $7+3(x+2)+4{{(x+2)}^{2}}-8{{(x+2)}^{3}}$ (Choice C) C $7+3(x+2)+2{{(x+2)}^{2}}-\frac{4}{3}{{(x+2)}^{3}}$ (Choice D) D $7+3(x-2)+2{{(x-2)}^{2}}-8{{(x-2)}^{3}}$ (Choice E) E $7+3(x+2)+2{{(x+2)}^{2}}-\frac{8}{3}{{(x+2)}^{3}}$
Explanation: We know the formula for the Taylor series centered at $~x=-2~$ for the function $~f\,$. $ f(-2)+f\,^\prime(-2)(x+2)+\frac{f\,^{\prime\prime}(-2)}{2!}{{(x+2)}^{2}}++...+\frac{{{f}^{(n)}}(-2)}{n!}{{(x+2)}^{n}}+...$ If we substitute the given function and derivative values, we obtain the following third-degree Taylor polynomial. $ T_3(x)=7+3(x+2)+\frac{4}{2!}{{(x+2)}^{2}}-\frac{8}{3!}{{(x+2)}^{3}}$ This simplifies to $ T_3(x)=7+3(x+2)+2{{(x+2)}^{2}}-\frac{4}{3}{{(x+2)}^{3}}\,$.